# Proof 8: The sum of squares

The formula for the sum of squares is:

$$1^2+2^2+\cdots+n^2=\frac16 n(n+1)(2n+1)$$

It is useful because it converts a row of confusingly many terms (therefore the dots $\cdots$) into a tangible number of terms that we can calculate.

If the formula works, then it must work for *any *number of terms. If one more term was added, the formula must therefore retain its form if it does work. We therefore add one extra term on the left-hand-side to see if the right-hand-side can be rearranged to the same form again. Such an extra term would be $(n+1)^2$, since the number $ n+1 $ is 1 larger than $n$:

$$\begin{align}

~& 1^2+2^2+\cdots+n^2 \quad\boldsymbol{+(n+1)^2}\\

~=&\underbrace{ \frac16 n(n+1)(2n+1)}_\text{replaced with the formula} \quad\boldsymbol{+(n+1)^2} \\

~=&(n+1)\left(\frac16 n(2n+1) +(n+1)\right) \\

~=&(n+1)\left(\frac16 n(2n+1) + \frac66 (n+1)\right) \\

~=&(n+1) \frac16 \left(n(2n+1) + 6 (n+1)\right) \\

~=&(n+1) \frac16 \left((2n^2+n) + (6n+6)\right) \\

~=&(n+1) \frac16 (2n^2+n + 6n+6) \\

~=&(n+1) \frac16 (2n^2+7n+6) \\

~=&(n+1) \frac16 (n+2)(2n+3) \\

~=&(n+1) \frac16 (n+1+1)(2n+2+1) \\

~=&\boldsymbol{(n+1)} \frac16 \left(\boldsymbol{(n+1)}+1\right)\left(2\boldsymbol{(n+1)}+1\right) \\

\end{align}$$

The expression here is now exactly the same as we started with, with a row of numbers with $n+1$ terms (it is the same formula $\frac16n(n+1)(2n+1)$ just with $n-1$ instead of $n$ to start with). Therefore, the formula works. $\quad_\blacksquare$

## Responses